Question: Complete the square to solve for $x$. $x^{2}+2x-48 = 0$
Begin by moving the constant term to the right side of the equation. $x^2 + 2x = 48$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $2$ , half of it would be $1$ , and squaring it gives us ${1}$ $x^2 + 2x { + 1} = 48 { + 1}$ We can now rewrite the left side of the equation as a squared term. $( x + 1 )^2 = 49$ Take the square root of both sides. $x + 1 = \pm7$ Isolate $x$ to find the solution(s). $x = -1\pm7$ So the solutions are: $x = 6 \text{ or } x = -8$ We already found the completed square: $( x + 1 )^2 = 49$